Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $a = \dfrac{k - 3}{3k - 15} \div \dfrac{k - 3}{k^2 - 11k + 30} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{k - 3}{3k - 15} \times \dfrac{k^2 - 11k + 30}{k - 3} $ First factor the quadratic. $a = \dfrac{k - 3}{3k - 15} \times \dfrac{(k - 5)(k - 6)}{k - 3} $ Then factor out any other terms. $a = \dfrac{k - 3}{3(k - 5)} \times \dfrac{(k - 5)(k - 6)}{k - 3} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (k - 3) \times (k - 5)(k - 6) } { 3(k - 5) \times (k - 3) } $ $a = \dfrac{ (k - 3)(k - 5)(k - 6)}{ 3(k - 5)(k - 3)} $ Notice that $(k - 3)$ and $(k - 5)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ \cancel{(k - 3)}(k - 5)(k - 6)}{ 3\cancel{(k - 5)}(k - 3)} $ We are dividing by $k - 5$ , so $k - 5 \neq 0$ Therefore, $k \neq 5$ $a = \dfrac{ \cancel{(k - 3)}\cancel{(k - 5)}(k - 6)}{ 3\cancel{(k - 5)}\cancel{(k - 3)}} $ We are dividing by $k - 3$ , so $k - 3 \neq 0$ Therefore, $k \neq 3$ $a = \dfrac{k - 6}{3} ; \space k \neq 5 ; \space k \neq 3 $